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Jimmy Metellus (extra credit)
For this reaction i mostly used wade ch. 10 and some of my knowledge. I took my best shot at it since it could give me some extra credit and study for the final exam.

(Updated)
First HBr was added to remove the double bond and add a bromine. Ether was then added so Br will attach to Mg. H3CH=O was then added so a oxygen will form. Then H3O+ was added so an H will be placed on the Oxygen. H2SO4 was then added so the hydroxyl would leave and cause a double bond to be formed. BR NBS cause two br to be put. Ether was added again so that Mg will attach to the two Br. 2H3 CH=O was added so two oxygen will be added. H3O+ was then added to put H on the two oxygen. H2SO4 was then added to remove the two OH and replace it wil double bonds.

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Jimmy: This synthetic strategy is really elegant. You cut the number of steps in half by building symmetrically from the inside out. It seems like a workable scheme to me. The only thing I would suggest that you add is a brief paragraph describing the steps. Also a few picky things like the Mg metal reagent in the grignard step.
Bruce Bondurant
Nice!

Padideh Rezabakhshpour(extra credit)


Padideh: I like your use of the wittig reaction. As with Jimmy's synthesis, you take advantage of the symmetry of the molecule to reduce the number of steps. The advantage of the wittig reaction is that it prevens rearrangement from happening.
Bruce Bondurant